A153677 Minimal exponents m such that the fractional part of (1024/1000)^m obtains a minimum (when starting with m=1).

1, 68, 142, 341, 395, 490, 585, 1164, 1707, 26366, 41358, 46074, 120805, 147332, 184259, 205661, 385710, 522271, 3418770, 3675376, 9424094

Offset: 1

Author

Hieronymus Fischer, Jan 06 2009

Keywords

• nonn
• more

Comments

Recursive definition: a(1)=1, a(n) is the least positive integer m such that the fractional part of (1024/1000)^m is less than the fractional part of (1024/1000)^k for all k, 1 <= k < m.

a(21) >= 4.5*10^6. - David A. Corneth, Mar 15 2019

a(22) > 10^7. Robert Price, Mar 16 2019

Examples

			a(2)=68, since fract((1024/1000)^68) = 0.016456..., but fract((1024/1000)^k) >= 0.024 for 1 <= k <= 67; thus fract((1024/1000)^68) < fract((1024/1000)^k) for 1 <= k < 68.
		

Crossrefs

Cf. A081464, A153669, A153681, A154130, A153685, A153693, A153701, A137994, A153717.

Programs

Mathematica

$MaxExtraPrecision = 10000;
p = .999;
Select[Range[1, 50000],
If[FractionalPart[(1024/1000)^#] < p,
p = FractionalPart[(1024/1000)^#]; True] &] (* Robert Price, Mar 15 2019 *)

PARI

upto(n) = my(res = List(), r = 1, p = 1); for(i=1, n, c = frac(p *= 1.024); if(cDavid A. Corneth, Mar 15 2019

Formula

Recursion: a(1):=1, a(k):=min{ m>1 | fract((1024/1000)^m) < fract((1024/1000)^a(k-1))}, where fract(x) = x-floor(x).

Extensions

a(18) from Robert Price, Mar 15 2019

a(19)-a(20) from David A. Corneth, Mar 15 2019

a(21) from Robert Price, Mar 16 2019